# Edit Distance LeetCode Solution

Here, We see Edit Distance LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character
Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

## Edit Distance Leetcode Solution C++

class Solution {
public:
int minDistance(string word1, string word2) {
int m=word1.length(),n=word2.length();

int dp[m+1][n+1];

for(int i=0;i<=m;i++)
dp[i][0]=i;

for(int i=0;i<=n;i++)
dp[0][i]=i;

for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(word1[i-1]==word2[j-1])
dp[i][j]=dp[i-1][j-1];
else{
dp[i][j] = 1+ min(dp[i][j - 1], min(dp[i - 1][j], dp[i - 1][j - 1]));
}
}
}
return dp[m][n];
}
};Code language: PHP (php)

## Edit Distance Leetcode Solution Java

class Solution {
public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null) {
throw new IllegalArgumentException("Input strings are null");
}
int insertCost = 1;
int deleteCost = 1;
int replaceCost = 1;
int l1 = word1.length();
int l2 = word2.length();
if (l1 == 0) {
return l2 * insertCost;
}
if (l2 == 0) {
return l1 * deleteCost;
}
if (l1 > l2) {
return minDistance(word2, word1);
}
int[] dp = new int[l1 + 1];
for (int i = 1; i <= l1; i++) {
dp[i] = dp[i - 1] + deleteCost;
}
for (int j = 1; j <= l2; j++) {
int prev = dp[0];
dp[0] += insertCost; // l1 is blank, Inserting l2 chars in l1.
char c2 = word2.charAt(j - 1);
for (int i = 1; i <= l1; i++) {
char c1 = word1.charAt(i - 1);
int temp = dp[i];
if (c1 == c2) {
dp[i] = prev;
} else {
dp[i] = Math.min(prev + replaceCost, Math.min(dp[i - 1] + deleteCost, dp[i] + insertCost));
}
prev = temp;
}
}

return dp[l1];
}
}Code language: JavaScript (javascript)

## Edit Distance Leetcode Solution JavaScript

/**
* @param {string} word1
* @param {string} word2
* @return {number}
*/
var minDistance = function(word1, word2) {
let dp = Array(word1.length+1).fill(null).map(()=>(Array(word2.length+1).fill(0)));

for (let i=0;i<dp.length;i++) {
dp[i][0] = i
}

for (let i=0;i<dp[0].length;i++) {
dp[0][i] = i
}

for (let i = 1;i<dp.length;i++) {
for (let j=1;j<dp[0].length;j++) {
dp[i][j] = Math.min(
dp[i-1][j]+1, // left
dp[i][j-1]+1, // right
dp[i-1][j-1] + (word1[i-1]!=word2[j-1]?1:0) // diagonal
);
}
}
return dp[dp.length-1][dp[0].length-1];
};Code language: JavaScript (javascript)

## Edit Distance Leetcode Solution Python

class Solution(object):
def minDistance(self, word1, word2):
h, w = len(word1)+1, len(word2)+1
pre = [i for i in range(w)]
for i in range(1, h):
cur = [i for _ in range(w)]
for j in range(1, w):
cur[j] = min(pre[j-1]+(word1[i-1] != word2[j-1]), pre[j]+1, cur[j-1]+1)
pre = cur
return pre[-1]

# O(m*n) space
def minDistance1(self, word1, word2):
h, w = len(word1)+1, len(word2)+1
dp = [[0 for _ in range(w)] for _ in range(h)]
for i in range(h):
dp[i][0] = i
for j in range(w):
dp[0][j] = j
for i in range(1, h):
for j in range(1, w):
dp[i][j] = min(dp[i-1][j-1]+(word1[i-1]!=word2[j-1]), dp[i-1][j]+1, dp[i][j-1]+1)
return dp[-1][-1]
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